| Message |
|
|
If  is 45 degrees then only the body moves horizontal otherwise not.
|
|
|
|
|
Always consider moving from higer potential to lower potential as negative sign and moving from lower potential to higher potential as positive sign. The same can be also considered reversed. But only one of them is to be considered.
|
|
|
|
(a) The current just after the connections are made is given by
i = (9 - 6) / 10 = 0.3 amp
(b) The current after completely charging is
i = (9 - 6) / 1 = 3 amp.
|
|
|
|
|
Nodal points method is same as Kirchoff's laws. Use Kirchoff's laws.
|
|
|
|
Method suggested by elessar_iitkgp is correct.
|
|
|
|
|
If the velocity of the ball when it strikes the incline is u, then after striking the incline its velocity remains same and makes an angle (alpha) w.r.t. the normal of the incline. Now use the equations for motion of a projectile on an incline plane.
|
|
|
|
|
Evaluation of the maximum ground acceleration distribution, seismic microzonation, is an essential procedure for prevention and mitigation of earthquake disasters. The local acceleration effects, which are peculiar phenomena during atomic scale deposition process, were investigated byMolecular Dynamics (MD) simulation. The values of local acceleration were distributed widely for various surface orientations. Deposited atoms were accelerated along the potential energy surface, and accelerated values were evidently dependent on the local configuration of the surface. In contrast, the local acceleration became negligibly small for clusters consisting of many atoms. Acceleration distribution is not neccessary for your syllabus.
|
|
|
|
|
This problem is from H.C.Verma. Consider the first surface as plane and the first image is at ny, where n is the R.I. of the water and y is the distance from the plane surface at which the object is placed. This image works as object for the curved surface and the object distance from the second surface is (ny + h). Using the thick lens formula we get the second image distance v as (1 / v) = (n - 1) / R - n / (ny + h) . Now replace v with -R. Because this image will work as object for the mirror and this object should appear to be at Center of Curvature. You will finaly get y as y = (R - h) / n.
|
|
|
|
Wave equations are the same for your syllabus. That is you deal only with sinosoidal waves. These equations look like this.
Y = A sin ( kx + wt + q), This equation can be transverse or longitudinal. This can be for light or sound. We cannot make out until some details are given.
Since your syllabus does not deal with damping, you need not to worry.
|
|
|
|
|
Now (x - 2u) / (y - 2u) = 2 . Solve for the requirement.
|
|
|
|
let the man velocity be v = u i + u j in vector form.
Let the wind velocity be w = x i + y j.
There fore the relative velocity = w - v = (x - u) i + (y - u) j
But the direction of the relative velocity is from north. Therefore x - u = 0. or x = u.
Now when the man velocity is doubled 2v = 2u i + 2u j.
therefore w - v = relative velocity.
|
|
|
|
|
substitute (EC - q) = u and you will find that dq = -du. rest is simple.
|
|
|
|
|
Dear debabratanag99, I have already given you the answer, that scattering effect is for particles smaller in size than the wavelength. The wavelength of light is approximately from 3500 A to 7500 A. Theoretical models suggest that the average cluster may encompass as many as 90 H2O molecules at 0°C, so that very cold water can be thought of as a collection of ever-changing ice-like structures. Therefore there is no point in discussing about scattering. This explanation is flawed for more than one reason. Fog droplets are, on
average, smaller than cloud droplets, but they still are huge compared with
the wavelengths of visible light. Thus scattering of such light by fog is
essentially wavelength independent. Unfortunately, many people learn
(without caveats) Rayleigh's scattering law and then assume that it applies
to everything. They did not learn that this law is limited to scatterers
small compared with the wavelength and at wavelengths far from strong
absorption.
|
|
|
|
Dear MRIITIAN007,
You should know that I am an expert in this field. So before you could say anything wrong about my answer you should veryfy 100 times what you are going to say. I do not claim I am the best. Atleast for this standard of questions I am well experienced.
I rightly agree with you that you have freedom of expression. But it should be to express your answer not to coment about others aspecially the experts.
Good luck.
GoIIT team member.
|
|
|
|
|
Tyndall effect is not for FOG.
|
|
|
|
|
I think the answers given above are enough to MR.IITIAN007. If not you can continue your questioning. I will further reply. Before commenting you must realize that you are still a student. Stundents should have good manners.
|
|
|
|
There is no good reason why fog lights are yellow. Here is an
excellent explanation provided by Professor Craig Bohren of Penn State
University:
"First I'll give you the wrong explanation, which you can find here and
there. It goes something like this. As everyone knows, scattering (by
anything!) is always greater at the shortwavelength end of the visible
spectrum than at the longwavelength end. Lord Rayleigh showed this, didn't
he? Thus to obtain the greatest penentration of light through fog, you
should use the longest wavelength possible. Red is obviously unsuitable
because it is used for stop lights. So you compromise and use yellow
instead.
This explanation is flawed for more than one reason. Fog droplets are, on
average, smaller than cloud droplets, but they still are huge compared with
the wavelengths of visible light. Thus scattering of such light by fog is
essentially wavelength independent. Unfortunately, many people learn
(without caveats) Rayleigh's scattering law and then assume that it applies
to everything. They did not learn that this law is limited to scatterers
small compared with the wavelength and at wavelengths far from strong
absorption.
The second flaw is that in order to get yellow light in the first place you
need a filter. Note that yellow fog lights were in use when the only
available headlights were incandescent lamps. If you place a filter over a
white headlight, you get less transmitted light, and there goes your
increased penetration down the drain.
There are two possible explanations for yellow fog lights. One is that the
first designers of such lights were mislead because they did not understand
the limitations of Rayleigh's scattering law and did not know the size
distribution of fog droplets. The other explanation is that someone deemed
it desirable to make fog lights yellow as a way of signalling to other
drivers that visibility is poor and thus caution is in order.
Designers of headlights have known for a long time that there is no magic
color that gives great penetration. I have an article from the Journal of
Scientific Instruments published in October 1938 (Vol. XV, pp. 317-322).
The article is by J. H. Nelson and is entitled "Optics of headlights". The
penultimate section in this paper is on "fog lamps". Nelson notes that
"there is almost complete agreement among designers of fog lamps, and this
agreement is in most cases extended to the colour of the light to be used.
Although there are still many lamps on the road using yellow light, it
seems to be becoming recognized that there is no filter, which, when placed
in front of a lamp, will improve the penetration power of that lamp."
This was written 61 years ago. Its author uses a few words ("seem",
"becoming recognized") indicating that perhaps at one time lamp designers
thought that yellow lights had greater penetrating power. And it may be
that because of this the first fog lamps were yellow. Once the practice of
making such lamps yellow began it just continued because of custom."
Also, take a look at the following web site:
http://www.gi.alaska.edu/ScienceForum/ASF5/593.html
Dr. Lawrence D. Woolf
General Atomics
|
|
|
|
Scattering of Light by small particles and molecules in the atmosphere Different from reflection, where radiation is deflected in one direction, some particles and molecules found in the atmosphere have the ability to scatter solar radiation in all directions. The particles/molecules which scatter light are called scatterers and can also include particulates made by human industry.  Selective scattering (or Rayleigh scattering) occurs when certain particles are more effective at scattering a particular wavelength of light. Air molecules, like oxygen and nitrogen for example, are small in size and thus more effective at scattering shorter wavelengths of light (blue and violet). The selective scattering by air molecules is responsible for producing our blue skies on a clear sunny day. Another type of scattering (called Mie Scattering) is responsible for the white appearance of clouds. Cloud droplets with a diameter of 20 micrometers or so are large enough to scatter all visible wavelengths more or less equally. This means that almost all of the light which enters clouds will be scattered. Because all wavelengths are scattered, clouds appear to be white.  When clouds become very deep, less and less of the incoming solar radiation makes it through to the bottom of the cloud, which gives these clouds a darker appearance
|
|
|
|
Generally to get into IITs main critirea is your self study. Offcourse any institute can guide you. It is not the institute but the decipline you meet. Do not join an institute which gives more teaching and less time for your self. Institutes like Chaitanya, Narayana are worst. These institutes teach you hours together and do not give you time for self study. For IIT self study is as important as teaching.
And regarding the result. Any institute that bosts of ranks should also give statistics about how many students they have given coaching and how many got ranks. If the statistics say it is 50% and above then it is good. Also these institutes have branches all over India. Therefore there is no fun in giving only ranks they got, but they should give what percentage of students got from their coaching. Why would an institute give concession in fee if they can deliver the quality? Why should they purchase ranks when they have quality?
Therefore it is the student who gets the rank. Work hard and choose any institute which delivers at the maximum 3 hours of coaching per day and also which gives students a chance to clarify doubts. Tests should be conducted properly and explanation to the questions in the test papers should be given after the test.
Now you know what is to be done.
|
|
|
|
|
It all depends on the requirement. May not be at IIT-B, but you may get else where.
|
|
|
|
E-Store
